How do you solve $6.2 e^(0.64t)=3e^(t+1)$ ?

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Answer 1 · 2017-05-05T10:15:42

See below.

$6.2 e^(0.64t)=3e^(t+1)$

so

$6.2/3=e^((1-0.64)t+1)$

or

$(1-0.64)t+1=log_e(6.2/3)$

then

$t=(log_e(6.2/3)-1)/(1-0.64) approx -0.761286$

How do you solve 6.2 e^(0.64t)=3e^(t+1)6.2 e^(0.64t)=3e^(t+1)?